s*****0
发帖数: 357 | 1 Essay写累了上来灌灌水，没想到脸红脖子粗的争了一场，多少有些意气了。忙了一天
，现在闲下心来好好说说这个话题。既然有网友抛砖引玉了，缄口不言也实在不够尊重
。另外版主塞了不少包子,也不能吃白食.
先说说常用的描述数据的几个概念.
1. random variation -- variability有两大类,一类是知道来源的,还有一类是
unexplained. 比如一部分variability是由treatment effect引起的,而剩下的大部分
variability却不知道缘故. 后者经常被归入random variation.
2. Mean -- 最常见的当然是arithmetic mean, 特殊情况下也会用geometric及
harmonic mean. 任何数据组都能计算mean, 但不是所有时候用mean都合适. 当碰到
extreme data values的时候(经常因为技术条件所限边缘值不太容易被精确测量),
median就会有很大的优越性, 因为median可以大幅降低个别extreme value的影响.
3. Median -- 概念不多提了, 优... 阅读全帖 |
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a****a
发帖数: 3411 | 2 QQ plot ---normality.我觉得正态分布的检测很麻烦，很多实验数据根本不够检测正态
性。用k-s test 估计大部分都是非正态。数据少了如果你测出来非正态，难道就是真的
吗？
levene's test ---eqality of variance
以前做过某细菌的双色荧光体系，每次在荧光显微镜下看到细菌的荧光颜色比例不均一
。但是比较对照组已经有明显差异。那时候的我的导师硬说我没有达到全部都是同色，
实验不理想。现在想想，这样的实验怎么能够做到。除非作弊。很多搞生物的没有概率
和统计的思维方式，可惜很多这样的人都是pi
y
经常有象我这样懂点统计又不是很懂的同志在用 t-test 的时候有些不放心。看了下面
这片文章后，我放心了。
Mihai Valcu & Cristina-Maria Valcu. 2010. Data transformation practices in
biomedical sciences. Nature Methods 8: 104–105.
文章写得很烂。但是对 t-test 的介绍还凑合。下面是我的一点理解。（英文是从文章... 阅读全帖 |
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J**Y
发帖数: 34 | 3 1. Heteroskedastic linear regression means that the random terms for each
observation are not iid. They are still indepenpendent but with different
variance. Does sigma here mean variance of random term? If it is true,
we just use ln(sigma)=x'b to estimate the var-cov matrix first, then use
it to do GLS. b is just parameter to decide variance. Taking log for sigma
here is just to transfrom dependent variable to normal distribution, or we can
say we assum sigma is log-normal distributed here. The |
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b*******n
发帖数: 51 | 4 我觉得你是在作FGLS，而不是GLS。GLS需要知道variance matrix的结构。如果是FGLS
，那么一般都是用searching algorithm。最简单的，先估计OLS，然后用OLS的
residual估计variance matrix。然后代到GLS估计的公式中，重新得到一组参数估计值
和residual。如果收敛了，就可以停下来了。如果不收敛，就需要重新search。另外，
看看Newey-West对variance的估计，细节我记不清楚了。你可以自己找。 |
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j**********y
发帖数: 1360 | 5 问一个统计的问题。
假设有两组variables，A组和B组，A组包含20个变量，B组也包含20个变量。
A1和B1是correlated, so is A2&B2, A3&B3，一直到A20到B20
如果把A组的每个变量的Variance当作一个新的变量，由VarA1, VarA2, 直到VarA20这
些数据点组成，
把B组的每个变量的Variance也当作一个新的变量，VarB1，VarB2，直到VarB20
那么这两个Variance变量是不是肯定就是相关的？ |
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D******n
发帖数: 2965 | 6 解个线性方程就性了。
1. Let X = (W, X_{sub}). 因为正态分布，对于每个W_j, 都可以写成一个关于X_sub
的线性方程加上一个独立的正态分布变量， i.e.,
W = b+ B * X_{sub} + V
where b is (n-m)*1 vector, B is (n-m)*m matrix, V independent of X_sub and
be joint normal with
zero mean, variance sigma_V -- (n-m)*(n-m) matrix.
use variance and covariance info of X to solve B, and sigma_V: covariance
between W and X_sub for B, and variance of W for sigma_V (the latter you
might not need given 2.)
(1) B* Sigma_{X_sub} = Cov (W, X_sub)
(2) B*Sigma_{X_sub} * B... 阅读全帖 |
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t****t
发帖数: 6806 | 7 给某些同学科普一下
白噪声是指一个*随机过程*, 而不是一些离散的sample. 你可以抽样,
但抽样后的点的variance也是无限的.
为什么? 功率谱和自相关函数是傅里叶变换对, 完全flat的功率谱就变
换为冲激的自相关函数. 自相关函数在0处的值就是功率, 对吧? 功率
就是variance, 对吧? 都是信号与系统的基本内容吧.
一般来说, 我们用的离散的n_i都是白噪声经过卷积/滤波得到的.
n_i=\int_0^T n(t)s(t)dt
时域卷积等于频域相乘. 那么噪声就变为了有限带宽/功率. 功率谱就变成了
S(F)的形状. 这样反变换回去, 自相关函数在0处才不为无限, 才有了有限variance.
这个卷积在某个时间处取抽样(0处? 取决于s的偏移), 才得到了n_i.
同时, 在不重叠的时间段卷积得到的n_i是independent的, 而分布又相同.
这才有了iid. |
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d****n
发帖数: 12461 | 8 是指DP算法么？这个具体是什么，能够大概解释一下么？
结论是对的。不过很荒谬啊。就是要100次全胜，基本没有胜算。
这样吧，投资组合里有mean-variance理论。考虑没一种投资策略对应mean-variance坐标
上的一个点，然后构成的集合的边界(mean大的方向)就是所有可行的策略。例如(mean,
var)=(1000,0)是一个。
最优策略一定在这个里面。只是不知道边界是什么样子的，虽然一定是mean关于variance
的递增函数。 |
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d****n
发帖数: 12461 | 9 是指DP算法么？这个具体是什么，能够大概解释一下么？
结论是对的。不过很荒谬啊。就是要100次全胜，基本没有胜算。
这样吧，投资组合里有mean-variance理论。考虑没一种投资策略对应mean-variance坐标
上的一个点，然后构成的集合的边界(mean大的方向)就是所有可行的策略。例如(mean,
var)=(1000,0)是一个。
最优策略一定在这个里面。只是不知道边界是什么样子的，虽然一定是mean关于variance
的递增函数。 |
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B*********h
发帖数: 800 | 10 ☆─────────────────────────────────────☆
ThisTiger (aTiger) 于 (Fri Mar 9 10:53:25 2007) 提到:
Find out the expectation and variance of these:
(1) integrate of dW(t) from 0 to T
(2) integrate of W(t)*dW(t) from W(0) to W(T)
(2) integrate of W(t)*dt from 0 to T
☆─────────────────────────────────────☆
ThisTiger (aTiger) 于 (Fri Mar 9 11:24:30 2007) 提到:
I guess you are on right track, however, to remind you, dW is of sqrt(t),
and its variance is of t, thus for WdW, the variance will be of t^ |
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i*****r
发帖数: 1302 | 11 有一列数据real variance挺大的,但real variance和bi-power variance很接近,所以
结果得出这个数据波动不算大,这算什么理论? bi-power是什么意思? |
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t********t
发帖数: 1264 | 12 volatility is not tradable, variance is.
refer to the variance swap model, constructed by Madan, Carr, etc.
The VIX index is actually calculated in the same way as the variance swap,
refer to CBOE's documents |
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s********7
发帖数: 52 | 13 当然不是保证赚钱
但是不是因为expected vol的问题。
你用option赌vol的最大特点是path dependent.
每2次delta hedge之间的PnL是0.5*$Gamma*(vol_realized^2-vol_imp^2)*T
如果某段t时刻，在你2次rebalance之间，realized比implied小，而此时刚好你的$
gamma又很大。而其他时刻虽然你realized比implied大，但是$gamma小，那么你到
maturity一算总的还是亏钱。
想消除这种path dependent的办法是用variance swap。这个比用option的好处就是和
股价无关。换而言之，不管是股价冲天还是股价暴跌，variance的$gamma都是constant
。实际上去年lehman破产，很多银行prop desk都long variance swap而不是去long
option来赌vol increase,就是这个道理。
hope it helps |
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t*****e
发帖数: 53 | 14 u = exp(sigma *sqrt(delta_t))
d = exp(-sigma *sqrt(delta_t))
p = (expt(r * delta_t) - d) / (u - d).
S_0 = 1;
At step n, whats the variance?
I follow the suggestion of this board and use the variance definition:
a node at nth is u^kd^(n-k). the probability is C_n_k * p^k * (1-p)^(n-k)
the expected stock price is exp(r * n *delta_t)
variance is sum over 0..n (C_n_k *p^k *(1-p)^(n-k) * (u^kd^(n-k)^2 - expt(-2
*n * delta_t) and I fursther simply this to:
(p*u^2 + (1-p) * d^2)^2 - expt(2*r*n*delta_t) |
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t*****e
发帖数: 53 | 15 you said the variance of the stock price return is not sigma^2*T, but the
variance of Ln(stock price return) is sigma^2 *T? right?
Can you prove the Ln(stock price return) is sigma^2 * T by variance
definition? |
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y******g
发帖数: 41 | 16 i see.
why would you say that calculating ACF has to have an assumption on the
variance?
btw, even if the conditional variance themselves are time-variant, the
unconditional variance in the long time are still constants by ARCH's
properties. |
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w**********y
发帖数: 1691 | 17 如果你要用bayesian来估计,那么MAP(maximum a posteriori estimation)应该是:
mean_p+sigma_p^2/(n*sigma_p^2+sigma^2)*mean
这里你的先验信息: sigma_p^2=25%, mean_p=12%
如果你的样本只有一个,那么n=1, mean=-2%
如果你假设variance不变,那么结果是:
12%-1%=11%
所以取决于你对variance的假设.
如果你假设posterior variance is unknown,那结果更复杂
文章的名字?? |
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a******u
发帖数: 66 | 18 T W_T得variance是T^3,然后减去一个variance为T^3/3,结果应该是个normal
distribution with variance=(2/3)T^3吧
dt |
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J**********g
发帖数: 213 | 19 The interview from MS was about fifty minutes, though it was supposed to be thirty
minutes long. My performance was...I don't know, not as good as expected. I
learned one thing from it, as said on Heard on the street:
ATQ: answer the quesiton.
I learned most thing by myself, so when I heard the question, it took time for me
to get what he wanted to know exactly. I asked him to repeat quesitons
several times. This is not good at all.
Again if this were my third or fourth interview company, I woul... 阅读全帖 |
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c****o
发帖数: 1280 | 20 you should first see that I(t) is normal distribution with mean 0 and
variance \int_0^t c(s)^2*exp(-2\int_s^t c(r)dr) ds, and in order to make it
goes to 0 with probability 1, I suggest variance goes to 0, and for the
variance, one such function I come up with is c(s) such that
(1) c(s)<=1
(2)c(\infty)=0
(3) \int_0^\infty c(s)=\infty
I might did the calculation wrong....... |
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c****o
发帖数: 1280 | 21 I will not disclose the name of the companys
1.Play a game. I am giving you $1, and the price of per share have mean $1,
variance 5 cents, how many share do I expected to get?(Hint: variance is
small quantity, use taylor expansion to 1/(1+x))
2.How to replicate binary option from call option, given the option prices
for ALL strike price, what is the value of the binary?
3.From 2, if the option pays a delta function, namely, pay 1 when s is
within epsilon neighborhood of K, price this option?
4.... 阅读全帖 |
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w******i
发帖数: 503 | 22 Thanks and congratulations, LZ. great post.
will not disclose the name of the companys
1.Play a game. I am giving you $1, and the price of per share have mean $1,
variance 5 cents, how many share do I expected to get?(Hint: variance is
small quantity, use taylor expansion to 1/(1+x))
2.How to replicate binary option from call option, given the option prices
for ALL strike price, what is the value of the binary?
3.From 2, if the option pays a delta function, namely, pay 1 when s is
within epsil... 阅读全帖 |
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l*********t
发帖数: 89 | 23 18. x,y~N(0,1), independent, what is E(x|x+y=1), what about variance?
Expectation为0.5， 但是variance没想出什么简单的方法，如果按照定义肯定可以做
，把x^2在conditional pdf下积分，再处理下即可..
不过variance有没有也像求expectation一样写个一行就能求出来的简单算法呢？
望大牛指点！多谢了~ |
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J*****n
发帖数: 4859 | 24
No, u r wrong, this system is essentially betting on the variance ratio of
stock return. So, in the random world, u can't make money, coz, random walk'
s variance ratio is 1. Ur example is actually mean-reverting, whose variance
ratio is less than 1. |
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t*******e
发帖数: 172 | 25 I can not help present the following idea.
It is easy to find two point distribution with probability 1/2 give us the
variance 1/4.
Then, we wanna prove it is an upper bound, I am not sure weather this method
was mentioned in this webpage.
Let us do the Monte Carlo Simulation, given a distribution p(x), we take n
samples from it. Then the variance is estimated by
(n(X_1^2+...+X_n^2)-(X_1+..+X_2)^2)/n^2, which is no more than 1/4, let n
tend to infinity, we know the variance no more than 1/4. Thi... 阅读全帖 |
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l*******1
发帖数: 113 | 26 Infinity.
overall gain = premium + hedging error
hedging error = -1/2*gamma*(realized variance - implied variance)
if realized variance is infinite, then hedging error is infinity...
lose insanely on the gamma. |
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y******i
发帖数: 199 | 27 我觉得这位的思路很好。
我是这样理解的。
首先，如果mean reverting（MR)和non-mean reverting(NMR)的SDE中的vol参数都给定
为相同值的话，那么期权价格肯定应该是相同的。
其次，如果我们考虑分别假设这两个模型，然后用同一个市场历史价格数据序列来
calibrate公式中的vol参数。
i)在有非常liquid的数据（近似连续的时间序列，并且bid-ask spread很小）的情况下
，由于drift term对quadratic variation是没有贡献的，所以calibrate出来的vol还
是一样的。所以期权价格还是应该相同。
ii)如果数据不是非常liquid，有离散性，比如只用日收盘价序列，那么drift term就
有影响了。这种情况下，我觉得calibrate出来的MR模型中的vol应该更大一些，因为其
variance中的一部分被mean reverting的效应抵消之后，net variance才和NMR模型中
vol贡献的variance相同。这种情况下，如果还是假设连续模型的话，则MR模型下的期
权价格还应该更高一些。
请... 阅读全帖 |
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z***e
发帖数: 5600 | 28 1. No, variance swap replication does not need to worry about shrinking of
T. Variance swap is the term variance throughout the period. Every day
passes by, there are fewer days and lower vega left. Vega goes to zero
instead of going to infinity when it gets close to expiration. Same with
the replicating options hence you do not need to rebalance because of change
of T.
5. Market practioners also use term structure trades as an alternative of
long/short vix. It tends to have lower beta. H... 阅读全帖 |
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L*******t
发帖数: 2385 | 29 Theoretically。。可以人为的加constraints，这个在理论上1992年就被Cvitanic和
Karatzas解决了，然后具体用起来可以假设一个Markov的structure用ＰＤＥ来解
然后Mean variance也可以加各种constraints。这样应该就不会有extreme了吧。此外
Mean variance要估计一个超大的Covariance matrix，然后参数不确定性的情况下，如
何用Mean Variance似乎也有结论。 |
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E***e
发帖数: 3430 | 30 手懶沒有具體算
W積分那個應該是寫成黎曼和就是一段段獨立的Gaussian increment
然後總和的variance就相當於每個variance加起來
最後取下極限
W^2積分那個類似道理但是好像變成Chi2了？
variance也是可以算的吧
感覺都不是很好玩就省下時間去鼓搗別的了 |
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v****x
发帖数: 980 | 31 In a local window, such as 5x5 window, calculate the variance of graylevel
for 9 pixels.then, sum or average all the variances, so called average local variance.
saying
clearly
way, |
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G****n
发帖数: 32 | 32 Can you please say more details? Do you mean I select a 5x5 window, then
calculate variance of first 3x3 pixels? then calculate all next? If that is
right, then I should get 9 variances from this 5x5 window. then average
these 9 values into one to compare?
Do you have any link for this method?
Thanks for help.
local variance. |
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v****x
发帖数: 980 | 33 sorry. it should be 25 pixels if you use 5*5 window, 9 pixels for 3x3
windows.
just calculate the variance of the graylevel for the 25 pixels and put this
variance in the center pixel. For any pixel, do the same calculation except
for edge pixels.
Then, just average all these variance. |
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s*****n
发帖数: 2174 | 34 Short answer: "the sample-size-5 variance" is they way to "compute the
theoretical variance".
Long answer:
It seems you have some confusions about basic concepts.
A distribution has an inherit variance and a lot of other attribute (such as
mean, skewness, etc), but neither of the distribution or the attributes of
the distribution can we see. What we can do is to obtain a sample from that
distribution, and use some function of the sample (a.k.a. statistics) to
guess the underlying distribution or |
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p********a
发帖数: 5352 | 35 ☆─────────────────────────────────────☆
PurpleShell (PurpleShell) 于 (Sun Jul 27 19:45:01 2008) 提到:
mean vector 知道，variance 矩阵知道，但不是正定的，如何构造具有多元正态分布
的随机向量？
☆─────────────────────────────────────☆
lisaxy (lisa) 于 (Mon Jul 28 01:12:46 2008) 提到:
variance 矩阵知道，但不是正定的?
variance 矩阵应该都是正定的吧。
☆─────────────────────────────────────☆
howmoney (多少钱) 于 (Mon Jul 28 10:13:12 2008) 提到:
In R:
library(mvtnorm)
then use rmvnorm(n, mean=c(), sigma=matrix())
☆─────────────────────────────────────☆
|
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h******a
发帖数: 4 | 36 在不同条件下 如何用EM Algorithm去估未知参数
1。Gaussian Mixture Model with common known variance
2。Gaussian Mixture Model with common unknown variance
3。Gaussian Mixture Model with different unknown variance |
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c******Z
发帖数: 160 | 37 需要用SAS9.1 （Analyst Application）做repeated measures， 但无论如何也不知道
该用什么covariance structure， 请牛人千万指点, 非常感谢！
Q1: Is the default covariance structure the variance components (VC) in
repeated measures?
Q2: Is VC actually the factorial ANOVA, with same variance but zero
covariance? Or,does VC contain different variance but zero covariance?
Q3: When trying VC, the p-value of “Null Model Likelihood Ratio Test” is
exactly 1. What does this mean and why?
Q4: When trying all OTHER covariance structures, the p- |
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o******6
发帖数: 538 | 38 ☆─────────────────────────────────────☆
davidmtl8 (davidmtl) 于 (Mon Feb 11 13:34:55 2008) 提到:
请问一个问题, two random variables x and y (do not know distributions of x
and y)
what is mean of (x/y), mean of ((x/y)2)
and variance of (x/y)
希望给些详细的说明,我是菜鸟一个.
谢谢
☆─────────────────────────────────────☆
davidmtl8 (davidmtl) 于 (Mon Feb 11 13:45:17 2008) 提到:
假设已知 mean of x and y, variance of x and y and covariance of (x,y)
ask for mean of (x/y), mean of ((x/y)2) and variance of (x/y)
☆────────────── |
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s*****n
发帖数: 2174 | 39 bandwidth 一般是local fitting的参数, 相当于在local多大的范围内进行regression
.
从bias-variance trade off的角度, bandwidth大, 则bias大, 但是variance小.
bandwidth小, 则bias小, variance大. Optimal bandwidth一般是通过某些预先设定的
loss function来找到的, 比如cross-validation.
从smoothing的角度, bandwidth相当于smoothing的强度, bandwidth越大, 结果越
smooth. |
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i*****r
发帖数: 1302 | 40 恕我无知,GLS是假设error的variance不是常数,不是error只应该有一个variance么?
OLS中 Y = X*B + e, e~N(mean, sigma), 假设有n个observation,难道这个sigma不
是n个error的variance?
不然GLS的error怎么会有个n*n的covariance matrix呢? |
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i*****r
发帖数: 1302 | 41 那如果是个matrix,对角线上的n个variance都是谁的variance? 假设n是天数,就是每天
的error都有一个variance? |
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h*******g
发帖数: 508 | 42 given a unknown gaussian distribution,
1) 如果有一批已知的采样值 from an unknown gaussian distribution， 我知道可以
根据 EM （estimated maximum likehood） 来推测出这个gaussian分布的最大可能性
2) 但如果这些采样个体的具体值未知， 只知道这些个体 被抽中的相对概率， 以及这
些个体的采样值的相对关系，
有什么算法可以去预测或者拟合出该gaussia的std or variance？
举个例子， 假设有一批数据是从一个未知的gaussian分布中提取的， 数据分为诺干个 subgroups, 已知所有这些数据的概率直，
还有在同一个subgroup里面,数据的相对关系已知（比如相邻样本的数值相差１）。
但这些数据的具体值未知，subgroups的相对关系也未知
如何去拟和出这个gaussian的std or variance？
由于数据的具体值未知，所以最后的gaussian分布的绝对位置可能是没法求的（mean）
， 但是variance或者std应该是可以拟合 |
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p*********8
发帖数: 1039 | 43 我试着答一下，我上次刚做过类似的作业
方法 delta method
1. let Z=X/Y,那Z 就是bivariate 的function
2. 用delta method 求出 Z 的mean, 和 variance， 这里的难点应该是variance, 它
是个矩阵的乘积，当然，结果是1*1，变成数值
3, U+/- 1.96* sqrt(variance)就是了 |
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g********d
发帖数: 2022 | 44 Required sample size increases as variance becomes larger.
If you are interested in proportions, 50% has the highest variance. In your
case, 20% has a variance of 0.2*0.8, while 10% has 0.1*0.9. Therefore, 20%
requires a larger sample size. |
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s***r
发帖数: 1121 | 45 Here is the original database
FirmID Year earnings Mean Variance
BBB 2004 E1 x1 std1
BBB 2005 E2 x2 std2
BBB 2006 E3 x3 std3
BBB 2007 E4 x4 std4
BBB 2008 E5 x5 std5
I want to randomly generate 50 numbers, normally distributed, with the mean/
variance as defined.
I want to create a new database like this:
FirmID Year earnings Mean Variance sim_num1 sim_num2.. sim_num50
BBB 2004 |
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P*******9
发帖数: 9700 | 46 i feel like you mixed the concepts of estimation and testing
for estimation you can talk about the efficiency/compare variance, but for t
est, you should compare power.
I think larger variance of test statistic doesn't mean anything
variance
it |
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d*******o
发帖数: 493 | 47 我觉得陈老师的想法很好。统计的预测理论都建立在两个基本点： least-square 基础上的
generalized linear model 和 nearest-neighbor的决策模型. least-square的优点是
low variance and high bias. nearest-neighbor 的优点是high variance and low
bias. 您的意思是想把两个结合起来，先segment再linearly fitting, 对吧？想
法很好啊。唯一的缺点就是这样的模型只能解释，不能用来预测，因为over-fitting会造成
同时的high variance and high bias. |
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h******3
发帖数: 190 | 48 The bonferroni method uses pooled variance. What if some pairs of sample
have much larger variance than others?
Do we have to use pooled variance? |
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d******e
发帖数: 7844 | 49 其实你完全不懂validation的意义。呵呵。
都是估计也是有很大不同的，你应该没听说过有一种东西叫做bias-variance trade
off。estimator的性能取决于估计的bias和估计的variance。
你以为估计是unbiased和low biased就是好的？用成百山千阶的多项式几乎可以完美
fit训练数据，bias极小，但variance极大，这种model的脑残程度跟大师和大师的
model真的有一拼啊。
呵呵。顺便多告诉你点知识，对于一个3维以上高斯分布，如果在意L2 risk下的性能的话
，最好的mean的estimator并不是unbiased的sample mean。不过这个结论对于你可能太
高端了，感兴趣的话就自己去看看吧。
http://en.wikipedia.org/wiki/James%E2%80%93Stein_estimator
没有意义的缘故。
用B去 |
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t****r
发帖数: 702 | 50 当然不是这样说说就有道理了，严格的证明也有很多人做啊，只是很多人都不会去看而
已。
而且统计很多做法，最重要的还是intuition，或者大师所津津乐道的哲学（逻辑）。你
所说的sufficient statistic，只是对poupulation mean 或者population variance来
说，所有样本中关于这两个参数的信息都被summarize到sample mean或者sample varia
nce中了。但是同时如果考虑到他们又是complete statistic，也就是说他们是包含所
有这些参数信息“最小的”statistics. 很显然，他们过滤掉了很多原来样本中有用的
信息。 最好的sufficient statistic就是样本本身哈。 很简单一个例子就是，我们要
刻画一个变量的分布，显然仅仅有mean和variance是不够的。除了normal 的
distribution之外，有相同mean和variance的分布有无穷多个。
再用tree classifer来做例子。如果我们可以试图定义一个“sufficient statistic”
来进行分类，那... 阅读全帖 |
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